Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 AMC 8 Problems/Problem 15"
(→‎Problem 15)
 
(22 intermediate revisions by 13 users not shown)
Line 1: Line 1:
−
==Problem==
+
==Problem 15==
 
Let the letters <math>F</math>,<math>L</math>,<math>Y</math>,<math>B</math>,<math>U</math>,<math>G</math> represent distinct digits. Suppose <math>\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}</math> is the greatest number that satisfies the equation
 
Let the letters <math>F</math>,<math>L</math>,<math>Y</math>,<math>B</math>,<math>U</math>,<math>G</math> represent distinct digits. Suppose <math>\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}</math> is the greatest number that satisfies the equation
  
Line 11: Line 11:
 
The highest that <math>FLYFLY</math> can be would have to be <math>124124</math>, and it cannot be higher than that because then it would exceed the <math>6</math>-digit limit set on <math>BUGBUG</math>.  
 
The highest that <math>FLYFLY</math> can be would have to be <math>124124</math>, and it cannot be higher than that because then it would exceed the <math>6</math>-digit limit set on <math>BUGBUG</math>.  
  
−
So, if we start at <math>124124\cdot8</math>, we get <math>992992</math>, which would be wrong because both <math>B</math> & <math>U</math> would be <math>9</math>, and the numbers cannot be repeated.  
+
So, if we start at <math>124124\cdot8</math>, we get <math>992992</math>, which would be wrong because both <math>B \& U</math> would be <math>9</math>, and the numbers cannot be repeated between different letters.  
  
−
If we move on to the next lowest, <math>123123</math>, and multiply by <math>8</math>, we get <math>984984</math>. All the digits are different, so <math>FLY+BUG</math> would be <math>123+984</math>, which is <math>1107</math>. So, the answer is <math>\boxed{\textbf{(C)}1107}</math>.  
+
If we move on to the next highest, <math>123123</math>, and multiply by <math>8</math>, we get <math>984984</math>. All the digits are different, so <math>FLY+BUG</math> would be <math>123+984</math>, which is <math>1107</math>. So, the answer is <math>\boxed{\textbf{(C)}1107}</math>.  
  
 
- Akhil Ravuri of John Adams Middle School
 
- Akhil Ravuri of John Adams Middle School
  
−
~ Aryan Varshney of John Adams Middle School (minor edits; props to Akhil Ravuri for the full solution :D)
+
 
 +
- Aryan Varshney of John Adams Middle School (minor edits... props to Akhil for the main/full answer :D)
  
 
~ cxsmi (minor formatting edits)
 
~ cxsmi (minor formatting edits)
Line 43: Line 44:
 
So, the correct answer is <math>\boxed{\textbf{(C)}\ 1107}</math>.
 
So, the correct answer is <math>\boxed{\textbf{(C)}\ 1107}</math>.
  
−
- C. Ren, Thomas Grover Middle School
+
- C. Ren
  
−
==Solution 3(Answer choices)==
+
==Solution 3 (Answer Choices)==
−
Note that <math>FLY+BUG = 9 \cdot FLY</math>. Thus, we can check the answer choices and find <math>FLY</math> through each of the answer choices, we find the 1107 works, so the answer is <math>\textbf{(C)}\ 1107</math>.
+
Note that <math>FLY+BUG = 9 \cdot FLY</math>. Thus, we can check the answer choices and find <math>FLY</math> through each of the answer choices, we find the 1107 works, so the answer is <math>\boxed{\textbf{(C)}\ 1107}</math>.
  
 
~andliu766
 
~andliu766
  
 
==Video Solution 1 by Math-X (First fully understand the problem!!!)==
 
==Video Solution 1 by Math-X (First fully understand the problem!!!)==
−
https://www.youtube.com/watch?v=JK4HWnqw-t0
+
https://youtu.be/BaE00H2SHQM?si=4za1LGPg_w2gwosm&t=3484
−
 
 
  
 
~Math-X
 
~Math-X
Line 64: Line 64:
 
==Video Solution 4 by SpreadTheMathLove==
 
==Video Solution 4 by SpreadTheMathLove==
 
https://www.youtube.com/watch?v=RRTxlduaDs8
 
https://www.youtube.com/watch?v=RRTxlduaDs8
 +
==Video Solution by NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=V-xN8Njd_Lc
 +
 +
~NiuniuMaths
  
 
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
  
 
https://www.youtube.com/watch?v=77UBBu1bKxk
 
https://www.youtube.com/watch?v=77UBBu1bKxk
 +
==Video Solution by Interstigation==
 +
https://youtu.be/ktzijuZtDas&t=1585
  
 
==See Also==
 
==See Also==
−
{{AMC8 box|year=2024|before=First Problem|num-a=15}}
+
{{AMC8 box|year=2024|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:46, 18 April 2024

Problem 15

Let the letters $F$,$L$,$Y$,$B$,$U$,$G$ represent distinct digits. Suppose $\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}$ is the greatest number that satisfies the equation

\[8\cdot\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}=\underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G}.\]

What is the value of $\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G}$?

$\textbf{(A)}\ 1089 \qquad \textbf{(B)}\ 1098 \qquad \textbf{(C)}\ 1107 \qquad \textbf{(D)}\ 1116 \qquad \textbf{(E)}\ 1125$

Solution 1

The highest that $FLYFLY$ can be would have to be $124124$, and it cannot be higher than that because then it would exceed the $6$-digit limit set on $BUGBUG$.

So, if we start at $124124\cdot8$, we get $992992$, which would be wrong because both $B \& U$ would be $9$, and the numbers cannot be repeated between different letters.

If we move on to the next highest, $123123$, and multiply by $8$, we get $984984$. All the digits are different, so $FLY+BUG$ would be $123+984$, which is $1107$. So, the answer is $\boxed{\textbf{(C)}1107}$.

- Akhil Ravuri of John Adams Middle School


- Aryan Varshney of John Adams Middle School (minor edits... props to Akhil for the main/full answer :D)

~ cxsmi (minor formatting edits)

Solution 2

Notice that $\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y} = 1000(\underline{F}~\underline{L}~\underline{Y}) + \underline{F}~\underline{L}~\underline{Y}$.

Likewise, $\underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G} = 1000(\underline{B}~\underline{U}~\underline{G}) + \underline{B}~\underline{U}~\underline{G}$.

Therefore, we have the following equation:

$8 \times 1001(\underline{F}~\underline{L}~\underline{Y}) = 1001(\underline{B}~\underline{U}~\underline{G})$.

Simplifying the equation gives

$8(\underline{F}~\underline{L}~\underline{Y}) = (\underline{B}~\underline{U}~\underline{G})$.

We can now use our equation to test each answer choice.

We have that $123123 \times 8 = 984984$, so we can find the sum:

$\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G} = 123 + 984 = 1107$.

So, the correct answer is $\boxed{\textbf{(C)}\ 1107}$.

- C. Ren

Solution 3 (Answer Choices)

Note that $FLY+BUG = 9 \cdot FLY$. Thus, we can check the answer choices and find $FLY$ through each of the answer choices, we find the 1107 works, so the answer is $\boxed{\textbf{(C)}\ 1107}$.

~andliu766

Video Solution 1 by Math-X (First fully understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=4za1LGPg_w2gwosm&t=3484

~Math-X

Video Solution 2 (easy to digest) by Power Solve

https://youtu.be/TKBVYMv__Bg

Video Solution 3 (2 minute solve, fast) by MegaMath

https://www.youtube.com/watch?v=QvJ1b0TzCTc

Video Solution 4 by SpreadTheMathLove

https://www.youtube.com/watch?v=RRTxlduaDs8

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=V-xN8Njd_Lc

~NiuniuMaths

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=77UBBu1bKxk

Video Solution by Interstigation

https://youtu.be/ktzijuZtDas&t=1585

See Also

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_8_Problems/Problem_15&oldid=218292"